Answer
$\dot{W}_{\max \text { out }}=157×10^6\text{ kW}$
Work Step by Step
First we determine the mole fraction of pure water in ocean water using Eqs. 13-4 and 13-5. Noting that $\mathrm{mf}_s=$ 0.025 and $\mathrm{mf}_w=1-\mathrm{mf}_s=0.975$, $$
\begin{aligned}
& M_{\mathrm{m}}=\frac{1}{\sum \frac{\mathrm{mf}_i}{M_i}}=\frac{1}{\frac{\mathrm{mf}_s}{M_s}+\frac{\mathrm{mf}_w}{M_w}}=\frac{1}{\frac{0.025}{58.44}+\frac{0.975}{18.0}}=18.32 \mathrm{~kg} / \mathrm{kmol} \\
& y_i=\mathrm{mf}_i \frac{M_m}{M_i} \rightarrow y_w=\mathrm{mf}_w \frac{M_m}{M_w}=(0.975) \frac{18.32 \mathrm{~kg} / \mathrm{kmol}}{18.0 \mathrm{~kg} / \mathrm{kmol}}=0.9922
\end{aligned}
$$ The maximum work output associated with mixing $1 \mathrm{~kg}$ of seawater (or the minimum work input required to produce $1 \mathrm{~kg}$ of freshwater from seawater) is
$$
w_{\max , \text { out }}=R_w T_0 \ln \left(1 / y_w\right)=(0.4615 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(288.15 \mathrm{~K}) \ln (1 / 0.9922)=1.046 \mathrm{~kJ} / \mathrm{kg} \text { fresh wate r }
$$ Therefore, $1.046 \mathrm{~kJ}$ of work can be produced as $1 \mathrm{~kg}$ of fresh water is mixed with seawater reversibly. Therefore, the power that can be generated as a river with a flow rate of $400,000 \mathrm{~m}^3 / \mathrm{s}$ mixes reversibly with seawater is $$
\dot{W}_{\max \text { out }}=\rho \dot{U}_{v_{\max \text { out }}}=\left(1000 \mathrm{~kg} / \mathrm{m}^3\right)\left(1.5 \times 10^5 \mathrm{~m}^3 / \mathrm{s}\right)(1.046 \mathrm{~kJ} / \mathrm{kg})\left(\frac{1 \mathrm{~kW}}{1 \mathrm{~kJ} / \mathrm{s}}\right)=157 \times 10^6 \mathrm{~kW}
$$
